ON THE GIRTHS OF REGULAR PLANAR GRAPHSMasakazu NihtvFujishiro High SchoolOne of the most fascinating yet mysterious classes of graphs are the cages.We introduce a planar version, classify them. and use this information topresent another proof of the fact that there are exactly five Platinic solids.We begin with a few definitions. The degree of a vertex v in a graph G isthe number of edges of G incident with v. A graph in which every vertex hasthe same degree is called a regular graph; if every vertex has degree k, thegraph is called a k-regular graph. The cardinality of the vertex set of G iscalled the order of G and is denoted by p. while the cardinality of its edge setis the size of G and is denoted by q. The length of the shortest cycle in a graphG that contains cycles is called the girth of G and is denoted by g(G) or g.Let us consider the k-regular graphs with girth g. The minimal order of ak-regular graph with girth g is denoted by f (k, g ) , and the k-regular graphs ofgirth g and order f(k, g) are called (k, g)-cages . For example, f ( 3,4) = 6and f (3,5) = <strong>10</strong>. The ( 3,4)-cage and (3, 5) -cage are unique and nonplanar,[I, 236-2391, 12.34-431. They are shown below.Fig. 1: (3,4) -cageFig. 2: @,5) -cageis not true.We will first determine the girths of all planar k-regular graphs for k = 4,5.THEOREM 1. If G is a connected planar +regular or 5-regular graph, theng(G) = 3.Proof. Let p, q, and s denote the order, size, and number of faces .of G.Then we have(1)andby Euler's formula.Let the distinct lengths of the boundaries of the faces of G be denoted byg = go,g,,-,g", (go sg,, i= L2,-. m).Suppose that there are s, faces with boundary of length g, . Then we have(3) E^=o s,g = 29.From (1) and (2) we have(4) y,":o s, = 2 +q(k- 2)/k,and from (3) we also have(5) O
Fig. 3: Girth 4 Fig. 4: Girth 5of the graphs in Figures 3 and 4 are 4 and 5, respectively.Since no planar k-regular graphs exist for k 2 6, we need consider only25 k < 5.Kk=4or5,theng=3 byTheorem 1. Whenk=2,a (2, g)-pageis the cycle whose length is g. If the graph G is a (5,3 )-page, then G mustsatisfip- q+s = 2, 3s 5 29, and 5p = 29.From this it is easy to see that the graph of Figure 5 is a (5,3 )-page, and thegraph of Figure 6 is a (4,3) -page.Fig. 5: A (5, 3) -pageFig. 6: A (4,3 ) -pageTherefore it remains only to investigate k = 3.TIIIOREM 2. If g(G) : 6, then a (3 .g)-page does not exist.lJrooJ Let G be a (3 ,g)-page of order 11 and si~e q. The for a (3, g) -page. we have g(q - p + 2 ) ,, a by Euler's fon~iula, Hence we obtainSince G is a 3-regular graph, ive also ha\e(8) 3p = 2q.Hence, with (7) and (8) we have(9)Thus g , 6.PROPOSITION. Let Gg be a (3, g) -page of order pf and six qg (g = 3, 4.5). Then (p3,q3) = (4.6). (p,.q,) = (8. 12). and (p5,q5) = (20,30).PruoJ We nu> deal onlj with g = 4. since the other cases are sin~ilar.Putting g = 4 In (7). we have q 2(p - 2). Therefore we obtain p, = 8 andq, = 12 by (8).Froni this proposition it is easj to check that the graphs of Figures 7. 3. and4 are a (3,3) -page. a (3 ,4)-page, and a (3,5)-page. respecti~el~.A regular pol>hedron is a poljhcdron whose faces are bounded b~congruent regular polygons and ~vhose polyhedral angles are congruent Thenevery regular polyhedron P is associated with a regular connected planar graphG(P) whose vertices and edges are the vertices and edges of 1'.If G(P) is a k-regular graphwith girth g, then the order of G(P)becomes n~inin~alsuch graphssince P is a regular polyhedron.So, G(P) becomes a (k, g) -pageWhen g ., 3. it is clear that thenumber of dimerent types of(k,g )-pages is only five by ourprevious results. This shows thatthe number of regular pol~licdra isat most five. On the other hand, Fig. 7: A (3.3) -pagewe can construct fi\e regularpolyhedra from the graphs of Figures 3-7 (the cube, dodecal~edron.~cosal~edron. octahedron. and tetral~edron. respecti~el~). We tl~erefore haie the\veII-known the or en^Theorcn~ 3 There are ewctl! fi\e regular polyhedra
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